Fields in which $x^3+x+1$ has a root such that no injective ring homomorphisms exist between fields Find 5 examples of fields in which the polynomial $f(x) = x^3 + x + 1$ has a root, and such that there are no injective ring homomorphisms between any two distinct fields among these 5 examples.
My attempt at a solution: $f(1)=3$, so $f(x)$ has a root in $\mathbb{Z}/3\mathbb{Z}$. $f(2)=11$, so $f(x)$ has a root in a root in $\mathbb{Z}/11\mathbb{Z}$. Similarly, $f(x)$ has a root in $\mathbb{Z}/23\mathbb{Z}$, $\mathbb{Z}/31\mathbb{Z}$, and $\mathbb{Z}/131\mathbb{Z}$. However, I don't know how to show that no injective homomorphisms exist between these fields (or if it's even true that there are no injective ring homomorphisms). Any help would be very much appreciated.
 A: An injective ring homomorphism would preserve the (additive) order of elements because it is also a homomorphism of groups if you just look at the additive structure. But then the order must divide the original group's order and the target group's order by Lagrange's theorem. Since they are co-prime, their only common factor is $1$, meaning no such homomorphism exists (the generators must be mapped to the identity).

Edit (op has not studied groups): The simple way is to note that $x+x+\ldots + x$ a total of $n$ times in $\Bbb Z/n$ is the $0$ element of the ring. So if you take $1\in\Bbb Z/n$ and look at $\phi(1)\in\Bbb Z/m$ we see that since $\phi$ is a homomorphism

$$0=\phi(n) = \phi(1+1+1+\ldots + 1) = \phi(1) +\ldots +\phi(1) = n\cdot\phi(1)$$

But then when $\gcd(m,n)=1$ there are $a,b$ so that $an+bm=1$, so $an\equiv 1\mod m$, i.e. $an\phi(1) \equiv \phi(1)\mod m$. That means that $n$ is not a zero divisor, i.e. if $nx=0$ it implies $x=0\mod m$. But this means $\phi(1)=0$ so $\phi(x) =0$ for all $x$, hence the homomorphism cannot be injective.
